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Good Bye Gravity!

Suppose you are standing on the surface of the earth and suddenly, the attraction due to gravity exerted by the earth vanishes. What would your position be (for example, you would fly up, fly tangentially or remain standing, etc) just at the instant after the attraction has vanished?
Also, what would be your position if the revolution of the earth is neglected?

Comments

Well, that does seem quite right but I was thinking that a person standing on the surface of the earth would tend to fly tangentially, in the case where the revolution of the earth is taken into consideration. Its something similar to the situation when you tie a stone at the end of a string and whirl it in a circle but if the string is suddenly cut, the stone flies away tangentially due to inertia of direction. What do you think about this possibility?

@Vaishnavi Gupta
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If you neglect everything like buildings,trees,air drag,etc. and assume the Earth to be a completely barren sphere, then -

Just before the gravitational force became 0, you were at rest w.r.t. the Earth. Thus w.r.t. a person at rest in space, you would be in a complicated motion-

circular due to rotation of the Earth and another bigger circular motion due to revolution of the Earth.

Thus when it becomes 0, you can't be at rest with Earth as the gravity, Normal and Friction all are 0.You would then possess the velocity which you had just before the gravity became 0. Now comes the fun part. If you were on the side facing the sun(i.e. if it was day-time when it happened) and your velocity's direction was in the path of the orbit of the Earth around the Sun(assuming the poor Sun still has its gravity), then the Earth will(or should) cancel out your path as it will move inwards(as it is following it's orbit) and so a Normal reaction force will act between you and the Earth which should propel you towards the sky(not straight upwards, but rather like the path of a reflected ray of light) and potentially towards the Sun! If it is night, then you should fly off tangentially.

This is all, of course, neglecting the gravitational pull of the Sun, the Moon and the other planets will have on you. If we take that into consideration, then it becomes much more complex (and intellectually enjoyable).

And lastly, I am not saying I am correct. Feel free to imagine better than me!

@Siddharth Brahmbhatt
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But wait, if you want to calculate the trajectory, use this.
Assuming Rotational speed=402m/s
As the rotation is 23.5degrees tilted, we should have a \(\Delta v_{horizontal}=\cos 23.5 \times 402\)

\(\Delta v_{vertical}=\sin 23.5 \times 402\)

\(e=\frac{r_{min}}{r_{max}}\)

\(a=r_{min}(1 + e)\)

\(v_{orbit}=\sqrt{\frac{(1+e) GM_{sun}}{r_{perihelion}}}\)

Ang.Distance from perihelion is \(\theta\)

\(\alpha=r_{min}(1+e)\)

\(r=\frac{\alpha}{1+e\cos\theta}\)

Also we want to find the speed of earth at that point. So, we can derive the equation from angular momentum' conservation.

now, \(\theta_{3}\) is the angle between periapsis of object.
when it is below 90, the apoapsis will raise more than periapsis. But, if it is above 90, it will raise the periapsis but \(\theta_{3}\leq180\)

So, \(\Delta periapsis=\cos\theta_{3}(\Delta v_{horizontal})\)

\(\Delta apoapsis=\sin\theta_{3}(\Delta v_{horizontal})\)

So, with that we can find that we are actually oscillating about the orbit of earth(but slightly higher one)

How's it? Hard one? Guys! these are only the beginning of Orbital Physics!